POJ 2993 Emag eht htiw Em Pleh解题报告

Description

This problem is a reverse case of the  problem 2996. You are given the output of the problem H and your task is to find the corresponding input.

Input

according to output of  problem 2996.

Output

according to input of  problem 2996.

Sample Input

White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4
Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6

Sample Output

+---+---+---+---+---+---+---+---+
|.r.|:::|.b.|:q:|.k.|:::|.n.|:r:|
+---+---+---+---+---+---+---+---+
|:p:|.p.|:p:|.p.|:p:|.p.|:::|.p.|
+---+---+---+---+---+---+---+---+
|...|:::|.n.|:::|...|:::|...|:p:|
+---+---+---+---+---+---+---+---+
|:::|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|...|:::|...|:::|.P.|:::|...|:::|
+---+---+---+---+---+---+---+---+
|:P:|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|.P.|:::|.P.|:P:|...|:P:|.P.|:P:|
+---+---+---+---+---+---+---+---+
|:R:|.N.|:B:|.Q.|:K:|.B.|:::|.R.|
+---+---+---+---+---+---+---+---+

这个题用了最笨的方法，结果RE了两次，调试了好久，原理就不说了，衰。。。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
using namespace std;
char a[18][34]={
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
{'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
{'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'}};

int main()
{
	int i,j;
	/*for(i=0;i<17;i++)
	{
		for(j=0;j<33;j++)
		{
			cout<<a[i][j];
		}
		cout<<endl;
	}*/
	char white[1000];
	char black[1000];
	int lenw,lenb;
	gets(white);
	gets(black);
	//cout<<white<<endl;
	//cout<<black<<endl;
	lenw=strlen(white);
	
	lenb=strlen(black);
/*	for(i=0;i<17;i++)
	{
		for(j=0;j<33;j++)
		{
			cout<<a[i][j];
		}
		cout<<endl;
	}*/
	i=7;
	for(i=7;i<lenw;i+=4)
	{
		if(white[i]>=65&&white[i]<=90)//判断第8个字符是不是大写字母。
		a[(9-(white[i+2]-'0'))*2-1][(white[i+1]-'a')*4+2]=white[i];
		else//不是就撤，表示后面只有P了
			break;
	}
	for(j=i;j<lenw;j+=3)
	{
		if(white[j]>=97&&white[j]<=122)//判断有没有小写字母
		a[(9-(white[j+1]-'0'))*2-1][(white[j]-'a')*4+2]='P';
		else//没有了就over了。
			break;
	}
	i=7;//同上哈，不罗嗦。
	for(i=7;i<lenb;i+=4)
	{
		if(black[i]>=65&&black[i]<=90)
		a[(17-(black[i+2]-'0')*2)][(black[i+1]-'a')*4+2]=black[i]+32;
		else 
			break;
	}
	for(j=i;j<lenb;j+=3)
	{
		if(black[j]>=97&&black[j]<=122)
		a[(17-(black[j+1]-'0')*2)][(black[j]-'a')*4+2]='p';
		else
			break;
	}
	for(i=0;i<17;i++)
	{
		for(j=0;j<33;j++)
		{
			cout<<a[i][j];
		}
		cout<<endl;
	}
	return 0;
}